\section{Introduction}\label{sec:intro} The four-vertex theorem says that the curvature of a simple closed plane curve that is not a circle has two local maxima separated by two local minima, with the values at both minima strictly smaller than those at both maxima. Gluck~\cite[pp.~295--296]{Gluck1971} proved the converse for continuous strictly positive curvature. Dahlberg~\cite[Theorem~1.1]{Dahlberg2005} proved it for smooth signed curvature, provided the two selected maxima are positive; see \cite{DGPV2007} for the history. The four-vertex theorem has many discrete versions \cite{Musin2004,OvsienkoTabachnikov2001,Sedykh1997,Tabachnikov2000,Dahlberg2008,Pak2010}. In the one studied here the curvature at a vertex of a polygon is the reciprocal radius of the circle through that vertex and its two neighbours. Let $P$ be a convex $n$-gon, $n\ge 4$, with vertices $x_1,\dots,x_n$ in cyclic order. Following Tabachnikov, call $P$ \emph{coherent} if for every $i$ the center of the circle through $x_{i-1},x_i,x_{i+1}$ lies inside the angle of $P$ at $x_i$. If $P$ is coherent and no four consecutive vertices lie on a circle, then the cyclic sequence $r_1,\dots,r_n$ of the radii of these circles has at least two local maxima and two local minima. This is due to Musin (see \cite{Tabachnikov2000,Musin2004}); Dahlberg~\cite{Dahlberg2008} proved a version for simple, locally regular polygons, not necessarily convex, extending Bilinski's result for convex polygons with obtuse angles~\cite{Bilinski1963}. In~\cite[Problem~2]{Tabachnikov2008} Tabachnikov posed the converse: a cyclic sequence $r_1,\dots,r_n$ with at least two local maxima and two local minima should be the radius sequence of a coherent convex polygon. Pak remarks that he is not aware of a nontrivial discrete version of the converse four-vertex theorem \cite[\S21.10, p.~206]{Pak2010}. We prove Tabachnikov's conjecture. Local extrema are counted as by Dahlberg \cite{Dahlberg2008}. A \emph{plateau maximum} of a cyclic sequence is a maximal run of equal consecutive terms, not the whole sequence, whose two outside neighbours are both strictly smaller. Plateau minima are defined in the same way. Two extrema are \emph{distinct} if their runs are disjoint. \begin{theorem}\label{thm:main} Let $n\ge 4$ and let $r_1,\dots,r_n>0$ be a cyclic sequence with at least two distinct plateau maxima and two distinct plateau minima. Then there is a simple, strictly convex, coherent polygon $x_1,\dots,x_n$, listed counterclockwise, whose circle through $x_{i-1},x_i,x_{i+1}$ has radius $r_i$ for every $i$. \end{theorem} % Lean: Gluck.Discrete.tabachnikov_converse_geometric (TabachnikovConverse.lean) Strict convexity means that, for each edge, all other vertices lie in the same open half-plane bounded by its supporting line; coherence places each circumcenter in the interior of the corresponding vertex angle. The same polygon has signed Menger curvature $1/r_i$. Moreover, whenever $r_i\ne r_{i+1}$, the four vertices $x_{i-1},x_i,x_{i+1},x_{i+2}$ do not lie on a common circle, independently of whether other adjacent radii are equal. % Lean: Gluck.Discrete.IsCoherentConvexPolygon.signedMengerR2_eq_inv_of_radii, % Gluck.Discrete.IsCoherentConvexPolygon.not_concyclic4_of_radius_ne % (TabachnikovCharacterization.lean) Two distinct plateau maxima already force two distinct plateau minima, and conversely (Lemma~\ref{lem:dfv-plateau}), so either half of the hypothesis suffices. \begin{figure}[t] \centering \includegraphics[width=0.7\textwidth]{tabachnikov/figures/example.pdf} \caption{A coherent convex octagon realizing the curvatures $(3,1.2,2.2,0.8,2.6,1,1.6,0.7)$, and its eight circles through consecutive vertex triples, whose centers lie inside the vertex angles. Computed numerically and checked to double precision.} \label{fig:example} \end{figure} Figure~\ref{fig:example} shows an example. The theorem is sharp: \begin{corollary}\label{cor:iff} Let $n\ge4$, and let $r_1,\dots,r_n$ be positive numbers that are not all equal. Then $r$ is the radius sequence of a coherent convex polygon if and only if $r$ has two distinct plateau maxima and two distinct plateau minima. \end{corollary} \begin{proof} One direction is Theorem~\ref{thm:main}. Conversely, a coherent convex polygon is simple, and its circumcenters lie in the closed vertex sectors, which is Dahlberg's local regularity \cite{Dahlberg2008}. It is not inscribed in a circle, since otherwise all the circles through consecutive triples would coincide and $r$ would be constant. Dahlberg's discrete four-vertex theorem \cite{Dahlberg2008}, whose extrema are plateau extrema with distinct runs, gives the two maxima and two minima. Only its convex case is needed here, since the polygon is strictly convex. The formalization includes this forward implication and the equivalence. % Lean: Gluck.Discrete.tabachnikov_radius_characterization \end{proof} \begin{remark}[The wording of the problem]\label{rem:wording} Read with strict extrema, $r_{i-1}r_{i+1}$, Tabachnikov's hypothesis is a special case of Theorem~\ref{thm:main}. Read with weak extrema, $r_{i-1}\le r_i\ge r_{i+1}$, the statement is false: $r=(2,1,1,1)$ has the weak local maxima $r_1,r_3$ and minima $r_2,r_4$ but one plateau maximum, so by Corollary~\ref{cor:iff} it is not a radius sequence. Constant sequences are excluded by our plateau convention and are realized by regular $n$-gons. Tabachnikov's non-concyclicity hypothesis holds for our polygon exactly when $r_i\ne r_{i+1}$ for all $i$: one implication was just noted, and if $r_i=r_{i+1}$ in a coherent convex polygon, the circles at $x_i$ and $x_{i+1}$ share the chord $x_ix_{i+1}$, the radius and the side of their centers, so they coincide. \end{remark} It is convenient to work with the curvatures $\kappa_i=1/r_i$. The hypothesis of Theorem~\ref{thm:main} is then equivalent to a \emph{strict four-vertex window}: indices $i_1